practice problems pending - tough

Q1) Let Γ be a semicircle with diameter AB. The point (C) lies on the diameter (AB) and points (E) and (D) lie on the arc (BA), with (E) between (B) and (D). Let the tangents to (\Gamma) at (D) and (E) meet at (F). Suppose that (\angle ACD = \angle ECB). Prove that (\angle EFD = \angle ACD + \angle ECB).

S1:
Let's quickly get an intuition about it.


If the point C is same as the center then CDFE is a cyclic quadrilateral since the angles at D and E are both 90. Angle at C is 180 - 2.theta so the angle at F is 2.theta which is what we need to show.
H.P.
So it works for the case when the point C is same as the center.

Now, let's try the general case.


Now ODFE is the cyclic quadrilateral.
=> Angle DFE = 180 - DOE
So proving that DFE = ACD + ECB = 2.theta is same as proving that:
DOE = 180 - 2.theta = DCE

But if we need to prove that DOE = DCE then we need to prove that E,D,C,O are concyclic since the chord DE is subtending the same angles at O,C.

Now look at the original problem statement and diagram.
A ray from D is hitting C and reflecting back at the same angle.
It's a strong hint that we need to use reflection.

Reflect D across AB to D'.

Angle D'CA = DCA by reflection symmetry.
Also D'CA = ECB as given.
=> D',C,E are colinear due to vertically opposite angles being equal.


Now, OE = OD' => OED' is an isosceles triangle.
=> Angle OD'E = OED' = OD'C = OEC
Also triangles OCD' and OCD are congruent by SSS.
=> Angle ODC = Angle OD'C

Combining them:
Angle OEC = ODC
=> The chord OC is subtending same angle at D,E
=> They are concyclic.
So we have proved that O,C,D,E are concyclic.
=> Angle DOE = DCE
=> Angle DFE = 180 - DOE = 180 - DCE
But DCE = 180 - 2.theta => DFE = 2.theta = DCA + ECB
H.P.

Q2.

Given a right-angled triangle (ABC) with hypotenuse (AB). A line passing through point (D), the midpoint of the hypotenuse (AB), intersects the lines (AC) and (BC) at points (P) and (Q), respectively. Let (M) be the midpoint of the segment (PQ). From point (R), which is symmetrical to point (D) with respect to point (M), a perpendicular (RF) is drawn to the hypotenuse (AB). Prove that (CM) is the bisector of angle (\angle FCD).

?


Comments

Post a Comment

Popular posts from this blog

IOQM 2024 Paper solutions (Done 1-21, 29)

Image
Q1 (number-theory). The smallest positive integer that does not divide 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 is: Solution : This product has all the integers from 1 to 9. So smallest integer not dividing it would be the next prime after 9. Which is 11. Answer: 11 Q2 (combinatorics): The number of four-digit odd numbers having digits 1, 2, 3, 4, each occurring exactly once, is: Solution: Last digit has to be 1 or 3: 2C1 Choosing remaining 3 digits: 3C3 Arranging those 3 digits: 3! Answer: 2C1*3C3*3! = 12 Q3 (number-theory): The number obtained by taking the last two digits of \(5^{2024}\) in the same order is: Solution: To get last 2 digits from number, divide by 100 and take the remainder. So we have to compute mod 100. 25 mod 100 = 25 25.5 mod 100 = 125 mod 100 = 25 5^3.5 mod 100 = 25.5 mod 100 = 25 5^4.5 mod 100 = 25.5 mod 100 = 25 ..... So: 5^2024 mod 100 = 25. Q4 (geometry): Let ABCD be a quadrilateral with \(\angle ADC = 70^\circ\), \(\angle ACD = 70^\circ\), \(\angle ACB = 10^\circ\...
Read more

Simon's factoring trick(complete the rectangle)

  "Complete the Rectangle" , also called Simon’s Favorite Factoring Trick , is a clever algebraic method for factoring expressions of the form: x y + a x + b y + c xy + ax + by + c Or more commonly, you'll see it used in a simpler form: x y + a x + b y + d xy + ax + by + d But especially when we’re given something like: x y + a x + b y + a b xy + ax + by + ab It becomes very easy to factor. Let me walk you through it step-by-step. 💡 The Key Idea We add and subtract a constant to turn the expression into a perfect rectangle (a.k.a. a factorable quadratic or product of binomials). The “complete the rectangle” version of this trick usually works best on expressions like: x y + a x + b y + a b xy + ax + by + ab We treat it like this: x y + a x + b y + a b = ( x + b ) ( y + a ) xy + ax + by + ab = (x + b)(y + a) ✅ Step-by-Step Example Factor: x y + 3 x + 2 y + 6 xy + 3x + 2y + 6 Step 1: Rearrange the terms: Group like this: x y + 3 x + 2 y + 6 xy + 3x + ...
Read more

IOQM 2023 solutions

Image
Q1. Let n be a positive integer such that 1 ≤ n ≤ 1000 . Let M n be the number of integers in the set X n = { 4 n + 1 , 4 n + 2 , … , 4 n + 1000 } . Let a = max ⁡ { M n : 1 ≤ n ≤ 1000 } , and b = min ⁡ { M n : 1 ≤ n ≤ 1000 } . a = \max\{M_n : 1 \leq n \leq 1000\}, \quad \text{and} \quad b = \min\{M_n : 1 \leq n \leq 1000\}. Find a − b a - b . Solution: Quick tip: If n^2 = k then (n+1)^2 = k + (2n + 1) We will use this here. Also as the numbers grow larger the gap between 2 perfect squares becomes less and less. For. e.g. there are 10 perfect squares between 1 and 100 but only 4 perfect squares between 101 and 200. So X1 = {sqrt(5) ... sqrt(1004)} will have the most number of perfect squares. While X1000 = {sqrt(4001)...sqrt(5000)} will have the least. In X1 the first integer square root is 3 and we know that 1024 is the square of 32 so the last integer will be 31. Total: 31 - 3 + 1 = 29 integers in X1. In X1000, We know that 64^2 = 4096 so 64 is the first integer. 70^2 = 4900 and ...
Read more